[math-fun] This will probably be incredibly stupid when we figure it out.
But that might be so easy. Plot says that f[InverseFunction[#1 - f[#1] &][t]] - InverseFunction[#1 - f[#1] &][t] == -t for the several different functions f that I tried. Here's f@t:=Sin[4t] <http://gosper.org/crazysinx-x.png>. (The gold overprinted the blue.) How can those two subexpressions plot so crazily, yet differ merely by the identity function? It's hard not to suspect Mathematica of malfeasance. Strangely methodical malfeasance. Regardless of the vibrations, does anyone understand this identity? Does anyone understand the vibrations? —rwg PS, did everyone (but me) get my (almost trivial) solution to c[0]:=0, c[n+1] = 2 + √c[n] ?
<< How can those two subexpressionsplot so crazily, yet differ merely by the identity function? >> I'm still struggling to understand the implied connection between these two clauses --- why should they not? WFL On 1/22/20, Bill Gosper <billgosper@gmail.com> wrote:
But that might be so easy. Plot says that f[InverseFunction[#1 - f[#1] &][t]] - InverseFunction[#1 - f[#1] &][t] == -t for the several different functions f that I tried. Here's f@t:=Sin[4t] <http://gosper.org/crazysinx-x.png>. (The gold overprinted the blue.) How can those two subexpressions plot so crazily, yet differ merely by the identity function? It's hard not to suspect Mathematica of malfeasance. Strangely methodical malfeasance. Regardless of the vibrations, does anyone understand this identity? Does anyone understand the vibrations? —rwg PS, did everyone (but me) get my (almost trivial) solution to c[0]:=0, c[n+1] = 2 + √c[n] ? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I just now received this, delayed exactly 4 days. Xmission.com is vile. —rwg On 2020-01-21 20:47, Bill Gosper wrote:
But that might be so easy. Plot says that f[InverseFunction[#1 - f[#1] &][t]] - InverseFunction[#1 - f[#1] &][t] == -t
for the several different functions f that I tried. Here's f@t:=Sin[4t] [1]. (The gold overprinted the blue.) How can those two subexpressions plot so crazily, yet differ merely by the identity function? It's hard not to suspect Mathematica of malfeasance. Strangely methodical malfeasance. Regardless of the vibrations, does anyone understand this identity? Does anyone understand the vibrations? —rwg PS, did everyone (but me) get my (almost trivial) solution to c[0]:=0, c[n+1] = 2 + √c[n] ?
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participants (3)
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Bill Gosper -
Fred Lunnon -
rwg